Proof of The Classes Cᵏ(U) Exist and Are Unique

Throughout, $\mathbb{R}^{U}$ is a set by The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set (ii) and its power set $\mathcal{P}(\mathbb{R}^{U})$ is a set by The Union Set and the Power Set of a Set (ii); subsets of $\mathbb{R}^{U}$ are formed by the set-builder of the setting, the conditions that define them (continuity on $U$, and the existence of partial derivatives on $U$ together with their membership in a given set) are predicative, and two sets with the same elements are equal.

We first fix the value at $0$. Let

$$c=\{h\in\mathbb{R}^{U}:\ h\text{ is continuous on }U\}.$$

This is a subset of $\mathbb{R}^{U}$, so $c\in\mathcal{P}(\mathbb{R}^{U})$; it is the set the first display requires for $\mathcal{C}^{0}(U)$.

Next we build the step of the recursion. Let $R$ be the class of pairs $((k,S),h)$ with $k\in\mathbb{N}_{0}$, $S\in\mathcal{P}(\mathbb{R}^{U})$ and $h\in\mathbb{R}^{U}$ such that, for every $i\in[n]$, the partial derivative $\partial_{i}h(x)$ exists for every $x\in U$ (Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ (ii)) and the partial derivative function $\partial_{i}h$ (Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ (iii)) lies in $S$. Applying Maps from Functional Predicative Relations, and Families of Sections (iii) to $R$, with $A=\mathbb{N}_{0}\times\mathcal{P}(\mathbb{R}^{U})$ and $b=\mathbb{R}^{U}$, we obtain a function $g:\mathbb{N}_{0}\times\mathcal{P}(\mathbb{R}^{U})\to\mathcal{P}(\mathbb{R}^{U})$ whose values are the sections of $R$:

$$h\in g(k,S)\iff h\in\mathbb{R}^{U}\ \text{and, for every }i\in[n],\ \partial_{i}h(x)\text{ exists for every }x\in U\text{ and }\partial_{i}h\in S .$$

Now we recurse. By The Recursion Theorem on Omega (i), applied with $\omega=\mathbb{N}_{0}$, $a=\mathcal{P}(\mathbb{R}^{U})$, the element $c$ and the function $g$, there is exactly one function $F:\mathbb{N}_{0}\to\mathcal{P}(\mathbb{R}^{U})$ with

$$F(0)=c\qquad\text{and}\qquad F(k+1)=g\bigl(k,F(k)\bigr)\quad\text{for every }k\in\mathbb{N}_{0},$$

where the successor of $k$ is $k+1$ by Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication (v). Put $\mathcal{C}^{k}(U)=F(k)$. Then $F(0)=c$ is the first display of the lemma, and by the description of $g(k,S)$ with $S=F(k)$, the elements of $F(k+1)$ are exactly the $h\in\mathbb{R}^{U}$ whose partial derivatives $\partial_{i}h$ exist on $U$ and lie in $\mathcal{C}^{k}(U)$, which is the second display. So a function with the required properties exists.

Finally, uniqueness. Let $F':\mathbb{N}_{0}\to\mathcal{P}(\mathbb{R}^{U})$ be any function satisfying both displays. The first display gives $F'(0)=c$. For each $k$, the second display and the description of $g(k,S)$ with $S=F'(k)$ show that $F'(k+1)$ and $g(k,F'(k))$ have the same elements, so $F'(k+1)=g(k,F'(k))$. Hence $F'$ satisfies the same recursion as $F$, and $F'=F$ by the uniqueness part of The Recursion Theorem on Omega (i). This proves (i).